Let $R$ be the region enclosed by the curves $y=\sqrt x$ and $y=\dfrac x3$. $y$ $x$ ${y=\sqrt x}$ ${y=\dfrac x3}$ $x=-1}$ $ 0$ $(9,3)$ $ R$ A solid is generated by rotating $R$ about the line $x=-1$. What is the volume of the solid? Give an exact answer in terms of $\pi$.
Explanation: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=\sqrt x}$ ${y=\dfrac x3}$ $x=-1}$ $ 0$ $(9,3)$ Let the thickness of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small thickness using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ This is called the washer method. What we now need is to figure out the expressions of $r_1(y)$ and $r_2(y)$, and the interval of integration. $r_1(y)$ is equal to the distance between the line $y=\dfrac x3$ and the line $x=-1$. To find it, we need to solve the equation for $x$ : $x=3y$ So, ${r_1(y)=3y+1}$. $r_2(y)$ is equal to the distance between the curve $y=\sqrt{x}$ and the line $x=-1$. To find it, we need to solve the equation for $x$ : $x=y^2$ So, ${r_2(y)=y^2+1}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [({r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi [({3y+1})^2-({y^2+1})^2] \\\\ &=\pi\left[ 9y^2+6y+1-(y^4+2y^2+1) \right] \\\\ &=\pi(-y^4+7y^2+6y) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=3$. So the interval of integration is $[0,3]$. Now we can express the definite integral in its entirety! $\begin{aligned} &\phantom{=}\int_0^3 \left[\pi (-y^4+7y^2+6y) \right] dy \\\\ &=\pi\int_0^3 (-y^4+7y^2+6y)dy \end{aligned}$ Let's evaluate the integral. $\pi\int_0^3 (-y^4+7y^2+6y)dy=\dfrac{207\pi}{5}$ In conclusion, the volume of the solid is $\dfrac{207\pi}{5}$.